A 5.0-kg crate is on an incline that makes an angle 30with the horizontal. If the coefficent of static friction is 0.5, the maximum force that can be applied parallel to the plane without moving the crate is:
A) 0
B) 3.3 N
C) 30 N
D) 46 N
E) 55 N

Question

A 5.0-kg crate is on an incline that makes an angle 30with the horizontal. If the coefficent of static friction is 0.5, the maximum force that can be applied parallel to the plane without moving the crate is:
A) 0 
B) 3.3 N 
C) 30 N 
D) 46 N 
E) 55 N

Quizplus · Accepted Answer

First, we need to find the force of gravity acting on the crate, which can be calculated as follows:

Fg = mg = (5.0 kg) x (9.81 m/s^2) = 49.05 N

Next, we need to find the maximum force that can be applied parallel to the plane without causing the crate to move. This is determined by the formula:

Ffmax = μsFn

where μs is the coefficient of static friction, and Fn is the normal force acting on the crate (i.e., the perpendicular component of the force of gravity).

Fn = Fgcos(30°) = (49.05 N)cos(30°) = 42.43 N

Ffmax = (0.5)(42.43 N) = 21.22 N

Therefore, the maximum force that can be applied parallel to the plane without causing the crate to move is 21.22 N. However, the question asks for the maximum force that can be applied, not just the maximum force of static friction.

To find this force, we need to use the formula:

Fmax = Ffmaxsin(30°) + Fgcos(30°)

Fmax = (21.22 N)sin(30°) + (49.05 N)cos(30°)

Fmax = 45.98 N

Therefore, the maximum force that can be applied parallel to the plane without causing the crate to move is 45.98 N. The closest answer choice is D, 46 N.

A 50-Kg Crate Is on an Incline That Makes an Angle